Question: Simplify the following expression and state the condition under which the simplification is valid. $r = \dfrac{q^2 - 64}{q + 8}$
Explanation: First factor the polynomial in the numerator. The numerator is in the form ${a^2} - {b^2}$ , which is a difference of two squares so we can factor it as $({a} + {b})({a} - {b})$ $ a = q$ $ b = \sqrt{64} = 8$ So we can rewrite the expression as: $r = \dfrac{({q} + {8})({q} {-8})} {q + 8} $ We can divide the numerator and denominator by $(q + 8)$ on condition that $q \neq -8$ Therefore $r = q - 8; q \neq -8$